3.1 \(\int \frac {\csc ^5(x)}{a+a \csc (x)} \, dx\)

Optimal. Leaf size=55 \[ -\frac {4 \cot ^3(x)}{3 a}-\frac {4 \cot (x)}{a}+\frac {3 \tanh ^{-1}(\cos (x))}{2 a}+\frac {\cot (x) \csc ^3(x)}{a \csc (x)+a}+\frac {3 \cot (x) \csc (x)}{2 a} \]

[Out]

3/2*arctanh(cos(x))/a-4*cot(x)/a-4/3*cot(x)^3/a+3/2*cot(x)*csc(x)/a+cot(x)*csc(x)^3/(a+a*csc(x))

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Rubi [A]  time = 0.07, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3818, 3787, 3768, 3770, 3767} \[ -\frac {4 \cot ^3(x)}{3 a}-\frac {4 \cot (x)}{a}+\frac {3 \tanh ^{-1}(\cos (x))}{2 a}+\frac {\cot (x) \csc ^3(x)}{a \csc (x)+a}+\frac {3 \cot (x) \csc (x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^5/(a + a*Csc[x]),x]

[Out]

(3*ArcTanh[Cos[x]])/(2*a) - (4*Cot[x])/a - (4*Cot[x]^3)/(3*a) + (3*Cot[x]*Csc[x])/(2*a) + (Cot[x]*Csc[x]^3)/(a
+ a*Csc[x])

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3818

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(d^2*Cot[e
+ f*x]*(d*Csc[e + f*x])^(n - 2))/(f*(a + b*Csc[e + f*x])), x] - Dist[d^2/(a*b), Int[(d*Csc[e + f*x])^(n - 2)*(
b*(n - 2) - a*(n - 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1]

Rubi steps

\begin {align*} \int \frac {\csc ^5(x)}{a+a \csc (x)} \, dx &=\frac {\cot (x) \csc ^3(x)}{a+a \csc (x)}-\frac {\int \csc ^3(x) (3 a-4 a \csc (x)) \, dx}{a^2}\\ &=\frac {\cot (x) \csc ^3(x)}{a+a \csc (x)}-\frac {3 \int \csc ^3(x) \, dx}{a}+\frac {4 \int \csc ^4(x) \, dx}{a}\\ &=\frac {3 \cot (x) \csc (x)}{2 a}+\frac {\cot (x) \csc ^3(x)}{a+a \csc (x)}-\frac {3 \int \csc (x) \, dx}{2 a}-\frac {4 \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (x)\right )}{a}\\ &=\frac {3 \tanh ^{-1}(\cos (x))}{2 a}-\frac {4 \cot (x)}{a}-\frac {4 \cot ^3(x)}{3 a}+\frac {3 \cot (x) \csc (x)}{2 a}+\frac {\cot (x) \csc ^3(x)}{a+a \csc (x)}\\ \end {align*}

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Mathematica [B]  time = 0.82, size = 113, normalized size = 2.05 \[ \frac {20 \tan \left (\frac {x}{2}\right )-20 \cot \left (\frac {x}{2}\right )+3 \csc ^2\left (\frac {x}{2}\right )-3 \sec ^2\left (\frac {x}{2}\right )-36 \log \left (\sin \left (\frac {x}{2}\right )\right )+36 \log \left (\cos \left (\frac {x}{2}\right )\right )+\frac {48 \sin \left (\frac {x}{2}\right )}{\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )}-\frac {1}{2} \sin (x) \csc ^4\left (\frac {x}{2}\right )+8 \sin ^4\left (\frac {x}{2}\right ) \csc ^3(x)}{24 a} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^5/(a + a*Csc[x]),x]

[Out]

(-20*Cot[x/2] + 3*Csc[x/2]^2 + 36*Log[Cos[x/2]] - 36*Log[Sin[x/2]] - 3*Sec[x/2]^2 + 8*Csc[x]^3*Sin[x/2]^4 + (4
8*Sin[x/2])/(Cos[x/2] + Sin[x/2]) - (Csc[x/2]^4*Sin[x])/2 + 20*Tan[x/2])/(24*a)

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fricas [B]  time = 0.54, size = 168, normalized size = 3.05 \[ \frac {32 \, \cos \relax (x)^{4} + 14 \, \cos \relax (x)^{3} - 48 \, \cos \relax (x)^{2} + 9 \, {\left (\cos \relax (x)^{4} - 2 \, \cos \relax (x)^{2} - {\left (\cos \relax (x)^{3} + \cos \relax (x)^{2} - \cos \relax (x) - 1\right )} \sin \relax (x) + 1\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) - 9 \, {\left (\cos \relax (x)^{4} - 2 \, \cos \relax (x)^{2} - {\left (\cos \relax (x)^{3} + \cos \relax (x)^{2} - \cos \relax (x) - 1\right )} \sin \relax (x) + 1\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + 2 \, {\left (16 \, \cos \relax (x)^{3} + 9 \, \cos \relax (x)^{2} - 15 \, \cos \relax (x) - 6\right )} \sin \relax (x) - 18 \, \cos \relax (x) + 12}{12 \, {\left (a \cos \relax (x)^{4} - 2 \, a \cos \relax (x)^{2} - {\left (a \cos \relax (x)^{3} + a \cos \relax (x)^{2} - a \cos \relax (x) - a\right )} \sin \relax (x) + a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^5/(a+a*csc(x)),x, algorithm="fricas")

[Out]

1/12*(32*cos(x)^4 + 14*cos(x)^3 - 48*cos(x)^2 + 9*(cos(x)^4 - 2*cos(x)^2 - (cos(x)^3 + cos(x)^2 - cos(x) - 1)*
sin(x) + 1)*log(1/2*cos(x) + 1/2) - 9*(cos(x)^4 - 2*cos(x)^2 - (cos(x)^3 + cos(x)^2 - cos(x) - 1)*sin(x) + 1)*
log(-1/2*cos(x) + 1/2) + 2*(16*cos(x)^3 + 9*cos(x)^2 - 15*cos(x) - 6)*sin(x) - 18*cos(x) + 12)/(a*cos(x)^4 - 2
*a*cos(x)^2 - (a*cos(x)^3 + a*cos(x)^2 - a*cos(x) - a)*sin(x) + a)

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giac [A]  time = 0.33, size = 96, normalized size = 1.75 \[ -\frac {3 \, \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{2 \, a} + \frac {a^{2} \tan \left (\frac {1}{2} \, x\right )^{3} - 3 \, a^{2} \tan \left (\frac {1}{2} \, x\right )^{2} + 21 \, a^{2} \tan \left (\frac {1}{2} \, x\right )}{24 \, a^{3}} - \frac {2}{a {\left (\tan \left (\frac {1}{2} \, x\right ) + 1\right )}} + \frac {66 \, \tan \left (\frac {1}{2} \, x\right )^{3} - 21 \, \tan \left (\frac {1}{2} \, x\right )^{2} + 3 \, \tan \left (\frac {1}{2} \, x\right ) - 1}{24 \, a \tan \left (\frac {1}{2} \, x\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^5/(a+a*csc(x)),x, algorithm="giac")

[Out]

-3/2*log(abs(tan(1/2*x)))/a + 1/24*(a^2*tan(1/2*x)^3 - 3*a^2*tan(1/2*x)^2 + 21*a^2*tan(1/2*x))/a^3 - 2/(a*(tan
(1/2*x) + 1)) + 1/24*(66*tan(1/2*x)^3 - 21*tan(1/2*x)^2 + 3*tan(1/2*x) - 1)/(a*tan(1/2*x)^3)

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maple [A]  time = 0.28, size = 89, normalized size = 1.62 \[ \frac {\tan ^{3}\left (\frac {x}{2}\right )}{24 a}-\frac {\tan ^{2}\left (\frac {x}{2}\right )}{8 a}+\frac {7 \tan \left (\frac {x}{2}\right )}{8 a}-\frac {1}{24 a \tan \left (\frac {x}{2}\right )^{3}}+\frac {1}{8 a \tan \left (\frac {x}{2}\right )^{2}}-\frac {7}{8 a \tan \left (\frac {x}{2}\right )}-\frac {3 \ln \left (\tan \left (\frac {x}{2}\right )\right )}{2 a}-\frac {2}{a \left (\tan \left (\frac {x}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)^5/(a+a*csc(x)),x)

[Out]

1/24/a*tan(1/2*x)^3-1/8/a*tan(1/2*x)^2+7/8/a*tan(1/2*x)-1/24/a/tan(1/2*x)^3+1/8/a/tan(1/2*x)^2-7/8/a/tan(1/2*x
)-3/2/a*ln(tan(1/2*x))-2/a/(tan(1/2*x)+1)

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maxima [B]  time = 0.32, size = 120, normalized size = 2.18 \[ \frac {\frac {21 \, \sin \relax (x)}{\cos \relax (x) + 1} - \frac {3 \, \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {\sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}}}{24 \, a} + \frac {\frac {2 \, \sin \relax (x)}{\cos \relax (x) + 1} - \frac {18 \, \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} - \frac {69 \, \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} - 1}{24 \, {\left (\frac {a \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {a \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}}\right )}} - \frac {3 \, \log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right )}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^5/(a+a*csc(x)),x, algorithm="maxima")

[Out]

1/24*(21*sin(x)/(cos(x) + 1) - 3*sin(x)^2/(cos(x) + 1)^2 + sin(x)^3/(cos(x) + 1)^3)/a + 1/24*(2*sin(x)/(cos(x)
+ 1) - 18*sin(x)^2/(cos(x) + 1)^2 - 69*sin(x)^3/(cos(x) + 1)^3 - 1)/(a*sin(x)^3/(cos(x) + 1)^3 + a*sin(x)^4/(
cos(x) + 1)^4) - 3/2*log(sin(x)/(cos(x) + 1))/a

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mupad [B]  time = 0.35, size = 89, normalized size = 1.62 \[ \frac {7\,\mathrm {tan}\left (\frac {x}{2}\right )}{8\,a}-\frac {23\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+6\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2-\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )}{3}+\frac {1}{3}}{8\,a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+8\,a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{8\,a}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{24\,a}-\frac {3\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{2\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)^5*(a + a/sin(x))),x)

[Out]

(7*tan(x/2))/(8*a) - (6*tan(x/2)^2 - (2*tan(x/2))/3 + 23*tan(x/2)^3 + 1/3)/(8*a*tan(x/2)^3 + 8*a*tan(x/2)^4) -
tan(x/2)^2/(8*a) + tan(x/2)^3/(24*a) - (3*log(tan(x/2)))/(2*a)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\csc ^{5}{\relax (x )}}{\csc {\relax (x )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)**5/(a+a*csc(x)),x)

[Out]

Integral(csc(x)**5/(csc(x) + 1), x)/a

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